Simplify and expand the following expression: $ \dfrac{4}{k + 7}+ \dfrac{3}{4k + 36}- \dfrac{4k}{k^2 + 16k + 63} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4k + 36} = \dfrac{3}{4(k + 9)}$ We can factor the quadratic in the third term: $ \dfrac{4k}{k^2 + 16k + 63} = \dfrac{4k}{(k + 7)(k + 9)}$ Now we have: $ \dfrac{4}{k + 7}+ \dfrac{3}{4(k + 9)}- \dfrac{4k}{(k + 7)(k + 9)} $ The least common multiple of the denominators is: $ (k + 7)(k + 9)$ In order to get the first term over $(k + 7)(k + 9)$ , multiply by $\dfrac{4(k + 9)}{4(k + 9)}$ $ \dfrac{4}{k + 7} \times \dfrac{4(k + 9)}{4(k + 9)} = \dfrac{16(k + 9)}{(k + 7)(k + 9)} $ In order to get the second term over $(k + 7)(k + 9)$ , multiply by $\dfrac{k + 7}{k + 7}$ $ \dfrac{3}{4(k + 9)} \times \dfrac{k + 7}{k + 7} = \dfrac{3(k + 7)}{(k + 7)(k + 9)} $ In order to get the third term over $(k + 7)(k + 9)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4k}{(k + 7)(k + 9)} \times \dfrac{4}{4} = \dfrac{16k}{(k + 7)(k + 9)} $ Now we have: $ \dfrac{16(k + 9)}{(k + 7)(k + 9)} + \dfrac{3(k + 7)}{(k + 7)(k + 9)} - \dfrac{16k}{(k + 7)(k + 9)} $ $ = \dfrac{ 16(k + 9) + 3(k + 7) - 16k} {(k + 7)(k + 9)} $ Expand: $ = \dfrac{16k + 144 + 3k + 21 - 16k}{4k^2 + 64k + 252} $ $ = \dfrac{3k + 165}{4k^2 + 64k + 252}$